Advertisement Remove all ads

Find the Vector Equation of the Plane that Contains the Lines → R = ( ˆ I + ˆ J ) + λ ( ˆ I + 2 ˆ J − ˆ K ) and the Point (–1, 3, –4). Also, Find the Length of the Perpendicular Drawn - Mathematics

Sum

Find the vector equation of the plane that contains the lines `vecr = (hat"i" + hat"j") + λ (hat"i" + 2hat"j" - hat"k")` and the point (–1, 3, –4). Also, find the length of the perpendicular drawn from the point (2, 1, 4) to the plane thus obtained.

Advertisement Remove all ads

Solution

Let the vector equation of the required plane be  `vec"r" . vec"n" = d`

The plane contains the line `vec"r" = hat"i" + hat"j" + λ (hat"i" + 2hat"j" - hat"k")`

Since the plane passes through point A and B. So `vec"n"` will be parallel to vector `vec"AB" xx (hat"i" + 2hat"j" - hat"k")`

`vec"AB" = vec"OB" -vec"OA"`

= `(-hat"i" + 3hat"j" - 4hat"k") (hat"i" + hat"j")`

= `-2hat"i" + 2hat"j" - 4hat"k"`

`vec"AB" xx (hat"i" + 2hat"j" - hat"k") = |(hat"i",hat"j",hat"k"),(1,2,-1),(-2,2, -4)|`

= `hat"i" (-8 + 2) - hat"j" (-4-2) + hat"k" (2+4)`

= `-6hat"i" + 6hat"j" + 6hat"k"`

which is a normal vector to the plane.

So the equation of plane will be `vec"r" . (-6hat"i" + 6hat"j" + 6hat"k") = d`

∴ it passes through (1, 1, 0) so `(hat"i" + hat"j"). (-6hat"i" + 6hat"j" + 6hat"k") = d or, d = 0`

equation of plane is `vecr . (-6hat"i" + 6hat"j" + 6hat"k") = 0`

`vecr (hat"i" - hat"j" -hat"k") = 0`

in Cartesian plane,

`(xhat"i" + yhat"j" + zhat"k") . (hat"i" - hat"j" -hat"k") = 0`

x -y - z = 0

So, the perpendicular distance of the plane from the point (2, 1, 4) is  ` = |(2 -1-4)/sqrt(1^2 + (-1)^2 + (-1)^2)| = |(-3)/sqrt(3)| = sqrt(3) "unit"`.

Concept: Vector and Cartesian Equation of a Plane
  Is there an error in this question or solution?
Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×